2 Answers
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Consider the force balance. The Earth exerts an upward force on the lake. Anything floating on the water is included in the weight of the lake. Since water is constant density, the upward force on the lake is a direct function of the water level - a higher level results in a higher pressure on the bottom, and a lower level results in a lower pressure on the bottom. Thus, no matter how you rearrange items supported by buoyancy on (or in) the lake, the water level will stay the same. An object that sinks touches the floor of the lake, so the prior argument no longer applies. The force balance dictates that the weight of the lake is less (as we consider the sunk object to no longer be a part of the hydrostatic body, since it's supported by a separate and unconnected normal force), and thus the water level will be less.

@adamdport The water level stays the same. The question of whether water itself is buoyant is a little more confused and frustrated. It both sinks and floats, it's the boundary between the two regions in the piecewise function that makes up the answer. Fortunately, however, the level is well defined in that case, it still stays the same. It would be different for oil or mercury.
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Alan RomingerOct 25 '12 at 14:34

Ah, yes, as Alan points out in his comment, in Basic buoyancy question: Man in a boat with a stone the equation for the volume of water displaced assumes the object is full submerged, and for a density less than water this wouldn't be the case. To briefly reprise the other question, when the object is in the boat the volume of water displaced is:

$$ V_{disp1} = \frac{M + m}{\rho_w} $$

and if the object is thrown from the boat and fully submerged the volume of water displaced is:

$$ V_{disp2} = \frac{M}{\rho_w} + \frac{m}{\rho_r} $$

where $\rho_r$ is the density of the object, so the change in volume displaced is:

$$ V_{disp1} - V_{disp2} = \frac{m}{\rho_w} - \frac{m}{\rho_r} $$

hence if $\rho_r$ > $\rho_w$ then $V_{disp1} > V_{disp2}$ and the water volume falls. But if the object has a density less than water, and therefore floats, the second equation changes to:

Not an exact duplicate because this question includes objects less than the density of water and the asking of that question does not.
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Alan RomingerOct 25 '12 at 13:52

Well the other question gives the answer in terms of $\rho_r$. Admittedly this is the density of a rock, but nothing in my answer constrains the value of $\rho_r$. It could be a rock made from pumice :-)
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John RennieOct 25 '12 at 13:55

@JohnRennie Correct me if I'm wrong, but I think that answer is actually wrong for $\rho_r<\rho_w$. Try it. I think it should be a piecewise function if applied to this question. It's just adding an adjustment for the case that it floats.
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Alan RomingerOct 25 '12 at 13:58

If the "object" is a astronomically giant air balloon (allowed in this question) then your answer predicts that water level is increase astronomically. This is only correct if God takes whatever the object is and jams it down into the lake against the force of buoyancy.
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Alan RomingerOct 25 '12 at 14:02

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@MelchiadeBedrosianBaol you should accept Alan's answer as I got the answer for $\rho_r < \rho_w$ wrong at the first attempt and wouldn't have noticed if he hadn't corrected me.
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John RennieOct 25 '12 at 14:37